OK, I'll work with that, BUT that is not how most people write the gamma density. I place beta in the denomiator.
The likelihood function is .

Thus with known we have suficient for .

And since we have UMVUE for .

I could have done this yesterday, but the mean of my gamma's is
and getting an unbiased estimator of is a lot harder in that case.
And I wasn't going to do this until I knew how you were writing your density.

Feb 28th 2010, 10:22 AM

ricer

Quote:

Originally Posted by matheagle

OK, I'll work with that, BUT that is not how most people write the gamma density. I place beta in the denomiator.
The likelihood function is .

Thus with known we have suficient for .

And since we have UMVUE for .

I could have done this yesterday, but the mean of my gamma's is
and getting an unbiased estimator of is a lot harder in that case.
And I wasn't going to do this until I knew how you were writing your density.

I met the same problem with this question.
if the the mean of gamma's is . How to get an unbiased estimator of ? Anyone can help? THANKS~(Thinking)

Apr 12th 2010, 08:06 PM

palabine

It can be shown that Sum(Xi) is not only sufficient but complete for beta (as w(beta) of exponential function contains an open set), so try 1/sum(xi) to estimate 1/beta. Lehmann-Scheffe tells us that an unbiased estimator that is a function of a complete statistic is the best unbiased estimator.

It can be shown that, assuming iid xi, Sum(xi) is distributed as Gamma(n*alpha,beta).

Given the above, let Y=(1/sum(xi)). Y is distributed as an inverted gamma(n*alpha, 1/beta) with mean=(1/beta)/(n*alpha-1). Thus, E(1/sum(xi)) = E(Y) = (1/beta)/(n*alpha-1), which is obviously a biased estimator of 1/beta.

Now let T = (n*alpha - 1)/sum(xi) be the unbiased estimator of 1/beta which is a function of a complete statistic. Thus, T is the best unbiased estimator of 1/beta, but it can be shown that it does not attain the lower Cramer-Rao bound.

But if you write down your pdf like this, doesn't this mean you have a gamma(a,1/b) instead of gamma(a,b)?

May 3rd 2011, 04:19 AM

Moo

Quote:

Originally Posted by mnazam

But if you write down your pdf like this, doesn't this mean you have a gamma(a,1/b) instead of gamma(a,b)?

There are 2 versions of the gamma distribution's pdf. You can have a look at the wikipedia page for the gamma distribution.

May 3rd 2011, 05:15 AM

mnazam

Quote:

Originally Posted by Moo

There are 2 versions of the gamma distribution's pdf. You can have a look at the wikipedia page for the gamma distribution.

I see. Thanks for the info. If the pdf can be written like this, it'd be so easy. Like we can show that it is an exponential family thus it is a complete sufficient estimator for beta. Then we can use MLE to get 1/beta(^). Then to show that 1/beta(^) is an unbiased estimator we calculate the mean of it and we get 1/beta. So 1/beta(^) in unbiased estimator.